User blog:P進大好きbot/Consistency of the Existence of a Linear Hierarchy Dominating All Functions
This is an English translation of my Japanese blog post submitted to a Japanese event. I denote by \(X\) the set of functions \(\mathbb{N} \to \mathbb{N}\), by \(X_0 \subset X\) the subset of strictly increasing functions, and by \(<_{\textrm{ed}}\) the strict partial ordering on \(X\) given as the eventual domination. Namely, for an \((f,g) \in X^2\), \(f <_{\textrm{ed}} g\) means that there exists some \(N \in \mathbb{N}\) such that \(n > N\) implies \(f(n) < g(n)\) for any \(n \in \mathbb{N}\). Let us consider the following axiom: I will show that Dominating Linear Hierarchy Axiom is consistent with \(\textrm{ZFC}\) set theory. I heard that the analogous proposition given by replacing \(X\) by the subset of computable functions \(\mathbb{N} \to \mathbb{N}\) was originally considered by Goedel. Namely, it is the question "Is there a linear hierarchy which dominates all computable functions?" It is still open, but is believed to be false by experts. For example, FGH along a totally ordered subset of Kleene's \(\mathcal{O}\) does not give a solution, because it consists of possibly uncomputable functions due to the uncomputability of fundamental sequences. I note that I have seen this problem or a quite similar problem somewhere, but I forgot where I saw it. In particular, this is not my original work. First of all, I recall an elementary fact: ; Proof. : If \(S\) is finite, then \(f\) can be chosen to be \(1 + \sum_{g \in S} g\). Therefore it suffices to show the assertion in the case where \(S\) is an infinite set. Take a bijective map \(\iota \colon \mathbb{N} \to S\). Denote by \(f \in X\) the function which assigns \(n + \sum_{i,j=0}^{n} \iota(i)(j)\) to each \(n \in \mathbb{N}\). For any \(n \in \mathbb{N}\), we have \(f(n) < f(n+1)\) by definition. Therefore \(f\) is strictly increasing. For any \(i \in \mathbb{N}\), \(n > i\) implies \(\iota(i)(n) < f(n)\) for any \(n \in \mathbb{N}\), and hence \(\iota(i) <_{\textrm{ed}} f\) holds. : □ Now it is time to show the main theorem. ; Proof. : By Cohen's theorem, it suffices to show Dominating Linear Hierarchy Axiom under \(\textrm{ZFC} + \textrm{CH}\). By \(\textrm{CH}\) and \(\# X = 2^{\aleph_0}\), there exists a bijective map \(\iota \colon \omega_1 \to X\). For each \(\alpha \in \omega_1\), we construct an order-preserving map \(F_{\alpha} \colon (\alpha+1,\in) \to (X_0,<_{\textrm{ed}})\) whose values are by transfinite induction on \(\alpha\). Put \(S_{\alpha} := \{\iota(\alpha)\} \cup \bigcup_{\beta \in \alpha} F_{\beta}(\beta)\). Since \(\alpha\) is countable, so is \(S_{\alpha}\). Therefore \(\{f \in X_0 \mid \forall g \in S_{\alpha}, g <_{\textrm{ed}} f\}\) is non-empty by the countable domination Lemma. By the bijectivity of \(\iota\) and the well-foundedness of \(\omega_1\), \(\{\delta \in \omega_1 \mid \iota(\delta) \in X_0 \land \forall g \in S_{\alpha}, g <_{\textrm{ed}} \iota(\delta)\}\) admits the least ordinal \(\delta_{\alpha}\). Define \(F_{\alpha} \colon (\alpha+1,\in) \to (X_0,<_{\textrm{ed}})\) as the map which assigns \(F_{\beta}(\beta)\) to each \(\beta \in \alpha\) and \(\iota(\delta_{\alpha})\) to \(\alpha\). By the construction, \((F_{\alpha})_{\alpha \in \omega_1}\) forms a compatible system of order preserving maps, and hence extends to a common order-preserving map \(F \colon (\omega_1,\in) \to (X_0,<_{\textrm{ed}})\). Denote by \(X_1 \subset X_0\) the image of \(F\). Since \(F\) is order-preserving and \(\omega_1\) is well-founded, \((X_1,<_{\textrm{ed}})\) is well-founded. Let \(f \in X\). By the construction of \(F\), we have \(f <_{\textrm{ed}} \delta_{\iota^{-1}(f)} = F(\iota^{-1}(f)) \in X_1\). Thus \(X_1\) is cofinal in \((X,<_{\textrm{ed}})\). : □ Therefore when we play uncomputable googology, we can use Dominating Linear Hierarchy under the Henkin extension of \(\textrm{ZFC}\) corresponding to Dominating Linear Hierarchy Axiom. (If we just work with \(\textrm{ZFC}\), then a Dominating Linear Hierarchy is not unique and hence is not specified. Or even it might not exist.) In this setting, creating a large function is essentially equivalent to creating a large countable ordinal, which is what many googologists are doing. On the other hand, if we play computable googology under \(\textrm{ZFC}\), it is not the case any more. Even if we define a large recursive ordinal, it does not directly give a computable large function. We need to encode it into an ordinal notation, and to define a recursive fundamental sequences for this purpose. Unfortunately, few googologists currently try to create an ordinal notation and a recursive system of fundamental sequences. Category:Blog posts